Question

All that is left in a packet of candy are 5 reds, 2 greens, and 2 blues. (a)What is the probability that a random drawing yields a green followed by a blue assuming that the first candy drawn is put back into the packet?

Answer 1

4/81

Explanation:

The candies in the packet are:

r = 5 (number of red candies)

g = 2 (number of green candies)

b = 2 (number of blue candies)

The total number of candies in the packet at the beginning is:

n = r + g + b = 5 + 2 +2 = 9

Therefore, at the first attempt, the probability of drawing a green candy is:

`p(g)=(g)/(n)=(2)/(9)`

Then, the first candy is placed back in the packet, so still

n = 9

Therefore, at the second attempt, the probability of drawing a blue candy is

`p(b)=(b)/(n)=(2)/(9)`

Therefore, the probability that a random drawing yields a green followed by a blue is:

`p(gb)=p(g)p(b)=(2)/(9)\cdot (2)/(9)=(4)/(81)`